How to find a particular Coefficient in a Binomial Expansion???

Question

Okay, I have to use the Binomial Theorem to find the indicated coefficient or term. The coefficient of x^0 in the expansion of (x- 1/x^2)^9. I did this problem and I end up with an x term in my answer (-1/x^18) which is completely different from what they have as an answer, a numerical answer. Please help me go through this step by step so I know what I did wrong and how to do this right...

I rewrote the expansion so that [x+ (-1/x^2)]^9

Then... I used the cute little shortcut so I wouldn't have to write out the ENTIRE expansion...

(n/ n-j) a^(n-j) x^j Note: There is no division sign here, it's just hard to separate n and n-j in this format.

n= 9
x= x
a= (-1/ x^2)
j=0

So plug and chug... (9 / 9-0)(-1/x^2)^9-0

= (9/9) (-1/x^18)x^0

= -1/ x^18

But that's not the right answer!!! HELP!!! What am I missing?

Huh?

:) I'll look over what you said again...

Answer 1

If the formula that is being raised had the form (x+a) with 'a' being constant, then j=0 would correspond to the term with x^0. But it doesn't, because 'a' is not constant. So you need to find the j that corresponds to the term with x^0.

Observe the following:

j=0 => term with exponent 0-2*9 = -18
j=1 => term with exponent 1-2*8 = -15
....
j=6 => term with exponent 6-2*3 = 0

So the answer you need is the term for j=6. That's where the 'x's will cancel out, giving you the answer corresponding to x^0.

If I'm not too sleepy, the answer is:

combinatorial(9;3) * (-1/x^2)^3 * x^6 = -84

Answer 2

I think maybe you are missing that you need to identify every term which results in x^0 then sum them.

if a=x and b=-1/x^2 then terms like b*a^2 result in x^0 so what terms in (a+b)^9 give x^0 ?

Best of Luck - Mike

Answer 3

(x-1/x^2)^9=[(x^3-1)/x^2]^9
=(x^3-1)^9/x^18

Sorry, maybe I didn't understand your question. How does one get a numerical number from the given expression which has a variable x? There seems to be no way to eliminate x in the given expression.

Answer 4

Your notation is very difficult to understand. This might be what you are looking for.

a^n-j . x^j
a = (-1/x²)
a = (- x^-²)
a^n-j . x^j In Binomial form this becomes:
(- x^-²)^9.x^0 + (- x^-²)^8.x^1 + (- x^-²)^7.x^2 + .........(- x^-²)^1.x^8 + (- x^-²)^0.x^9
Now check the question for a value of x, and put it's value in and calculate the result. It might be a good idea to stop taking short cuts for the time being.

Answer 5

hey the coefficients for the last term that is x^0 term have to be 1
becoz the last term is 9c9.x^0.(-1/x^2)^9
therefore i guess your answer is right

Answer 6

nicely, you could search for that contained in the nth row, mth column it really is C(n,m) = n! / [(n-m)! m!] the position n! is n factorial ( n*(n-a million)*(n-2)...3*2*a million ) (there's a 0th row and 0th column contained in the triangle. C(n,0) = a million) each and each variety contained in the triangle is the sum of both numbers above it. this would nicely be shown algebraicly, C(n,m) + C(n, m+a million) = n! / [(n-m)! m!] + n! / [(n-m-a million)! (m+a million)!] = { n! / [(n-m-a million)! m!] } * { a million/(n-m) + a million/(m+a million) } = { n! / [(n-m-a million)! m!] } * { (m+a million+n-m) / [(n-m)(m+a million)] } = { n! / [(n-m-a million)! m!] } * { (n+a million) / [(n-m)(m+a million)] } = (n+a million)! /[(n-m)! (m+a million)!] = C(n+a million, m+a million) to work out the way it truly is ideal to the binomial boost, evaluate (x+a million)^n shall we are saying that this equals a(0)*(x^n)*(a million^0) + a(a million)(x^(n-a million))*(a million^a million) + ... + a(n) , the position a(ok) = C(n,ok) If we multiply this through yet another (x+a million) , we get: (x+a million)^(n) * (x+a million) = a(0)*(x^(n+a million))*(a million^0) + a(a million)(x^n)*(a million^a million) + ... + a(n)x + a(0)*x^n*(a million^0) + a(a million)*x^(n-a million)*(a million^a million) + ... + a(n) + a million = a(0) x^(n+a million) + [a(a million) + a(0)]x^n + [a(2) + a(a million)]x^(n-a million) + ... + a million = (x+a million)^(n+a million) So each and anytime period is a(ok) + a(ok-a million), in different words, the sum of both words above it in Pascal's triangle.